Past exam of the mathematics course of the University of Cambridge / 2007 / ib / Paper 1 / 10G (Groups, Rings and Modules)

The Fundamental theorem of finitely generated abelian groups says that every finitely generated abelian group is isomorphic toThe rank of an abelian group $r$ and the invariant factors $d_j$ are uniquely determined, with the finite summand allowed to be absent. Equivalently, one can decompose the torsion subgroup into cyclic groups of prime-power orders. The infinite cyclic part and torsion part must both be included in the statement.

A nonzero polynomial of degree $n$ over a field has at most $n$ distinct roots in that field. For a proof, the factor theorem gives $(t-a)\mid f(t)$ whenever $f(a)=0$. Distinct linear factors are coprime in $K[t]$, a unique factorization domain, so for distinct roots $a_1,\dots ,a_k$ their product divides $f$. Its degree is $k$, giving $k\le n$. One may equivalently remove one linear factor at a time and argue by induction.

If the field has at least $n$ elements, this bound is attained by a product of $n$ distinct linear factors. For a fixed finite field with fewer than $n$ elements, the largest possible number of distinct roots is $\min (n,|K|)$ instead; multiplicities do not create more distinct roots. The zero polynomial is excluded since it has no finite degree.

Let $G$ be a finite subgroup of $K^{\times }$. It is abelian since multiplication in the field commutes. The structure theorem therefore writes $G$ as a product of cyclic groups with invariant factors $d_1\mid \cdots \mid d_s$, omitting trivial factors. Every element satisfies $g^{d_s}=1$, so $G$ consists of roots of $t^{d_s}-1$. The root bound from part (ii) giveswhile the decomposition gives $|G|={\prod }_j{d}_j\ge d_s$. Equality forces every factor except the last to be trivial. Hence $G$ is cyclic. The trivial subgroup is cyclic as well. This is the finite multiplicative subgroup of a field is cyclic theorem, proved here using the preceding two parts.

The infinite case is false. In ${\mathbb{C}}^{\times }$, the group of all roots of unity is infinite, but every element has finite order. If it were cyclic, its generator would have finite order and the whole group would be finite, a contradiction.